A steam enters in an insulated nozzle with a velocity of 40 m-s- The initial pressure and temperature of the steam is 6-87 bar and 225-C- It leaves at a pressure of 2-1 bar and a velocity of 450 m-s- The final enthalpy of the steam isAt 6-87 bar and 225-C-h- 2850 kJ-kg

The steady flow energy equation, h_1 +c_1^22+gz_1+Q = h_2+c_2^22+gz_2+W ∵ W = 0 , Q = 0, g(z_2 z_1) =0 h_2 h_1 =c_1^2 c_1^22 h_2 2850= (40)^2 (450)^22× 10^3 ...

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A steam enter...

Question

A steam enters in an insulated nozzle with a velocity of 40 m/s. The initial pressure and temperature of the steam is 6.87 bar and $225^{\circ} C .$ It leaves at a pressure of 2.1 bar and a velocity of 450 m/s. The final enthalpy of the steam is
(At 6.87 bar and $225^{\circ} C, h = 2850 k J / k g$ )

3000 kJ/kg

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2749.55 kJ/kg

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2250 kJ/kg

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2698 kJ/kg

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Solution

The correct option is B 2749.55 kJ/kg

$The steady flow energy equation, h_{1} + \frac{c_{1}^{2}}{2} + g z_{1} + Q = h_{2} + \frac{c_{2}^{2}}{2} + g z_{2} + W ∵ W = 0, Q = 0, g (z_{2} - z_{1}) = 0 h_{2} - h_{1} = \frac{c_{1}^{2} - c_{1}^{2}}{2} h_{2} - 2850 = \frac{(40)^{2} - (450)^{2}}{2 \times 10^{3}} \Rightarrow h_{2} = 2749. 55 k J / k g$

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A steam enters in an insulated nozzle with a velocity of 40 m/s. The initial pressure and temperature of the steam is 6.87 bar and 225∘C. It leaves at a pressure of 2.1 bar and a velocity of 450 m/s. The final enthalpy of the steam is (At 6.87 bar and 225∘C,h=2850 kJ/kg)

The correct option is B 2749.55 kJ/kg The steady flow energy equation,h1+c212+gz1+Q=h2+c222+gz2+W∵W=0,Q=0,g(z2−z1)=0h2−h1=c21−c212h2−2850=(40)2−(450)22×103⇒h2=2749.55 kJ/kg